Answer
$$\frac{3}{8}$$
Work Step by Step
$$\eqalign{
& \int_{ - \ln 2}^0 {{e^{2w}}} dw \cr
& {\text{use substitution}}{\text{: }} \cr
& {\text{ }}u = 2w,{\text{ so that }}\frac{{du}}{{dw}} = 2,\,\,\,dw = \frac{{du}}{2} \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}w = 0,{\text{ }}u = 2\left( 0 \right) = 0 \cr
& \,\,\,\,\,\,{\text{If }}w = - \ln 2,{\text{ }}u = 2\left( { - \ln 2} \right) = - 2\ln 2 \cr
& {\text{Then}} \cr
& \int_{ - \ln 2}^0 {{e^{2w}}} dw = \int_{ - 2\ln 2}^0 {{e^u}} \left( {\frac{{du}}{2}} \right) \cr
& = \frac{1}{2}\int_{ - 2\ln 2}^0 {{e^u}} du \cr
& {\text{integrate}} \cr
& = \frac{1}{2}\left( {{e^u}} \right)_{ - 2\ln 2}^0 \cr
& {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \frac{1}{2}\left( {{e^0} - {e^{ - 2\ln 2}}} \right) \cr
& {\text{simplifying, we get:}} \cr
& = \frac{1}{2}\left( {1 - \frac{1}{4}} \right) \cr
& = \frac{3}{8} \cr} $$
